IB DP Chemistry - R2.3.7 Gibbs free energy and equilibrium- Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – R2.3.7 Gibbs free energy and equilibrium – Study Notes – New Syllabus
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Reactivity 2.3.7 – Equilibrium Constant and Gibbs Free Energy Change
Reactivity 2.3.7 – Equilibrium Constant and Gibbs Free Energy Change
The position of equilibrium can be measured using either:
- The equilibrium constant \( K \), which describes the extent to which products or reactants are favored at equilibrium.
- The standard Gibbs free energy change \( \Delta G^\circ \), which indicates the spontaneity and energy change associated with a chemical reaction under standard conditions (1 bar, 298 K, 1 mol/L).
1. The Equation Relating \( \Delta G^\circ \) and \( K \)
The standard Gibbs free energy change is related to the equilibrium constant by the equation:
\( \Delta G^\circ = -RT \ln K \)
- \( \Delta G^\circ \) = standard Gibbs free energy change (J/mol)
- \( R \) = universal gas constant = 8.314 J mol–1 K–1
- \( T \) = temperature in kelvin (K)
- \( K \) = equilibrium constant (unitless)
Rearranged to solve for \( K \):
\( K = e^{-\Delta G^\circ / RT} \)
2. Interpreting the Relationship
| \( \Delta G^\circ \) | \( K \) | Interpretation |
|---|---|---|
| \( \Delta G^\circ < 0 \) | \( K > 1 \) | Forward reaction favored; products predominate at equilibrium. |
| \( \Delta G^\circ = 0 \) | \( K = 1 \) | Reactants and products equally favored at equilibrium. |
| \( \Delta G^\circ > 0 \) | \( K < 1 \) | Reverse reaction favored; reactants predominate at equilibrium. |
Important: This equation only applies under standard conditions. For non-standard conditions, use: \( \Delta G = \Delta G^\circ + RT \ln Q \)
Example
The standard Gibbs free energy change for a reaction is \( \Delta G^\circ = -34.2 \, \text{kJ/mol} \) at 298 K.
Calculate the equilibrium constant \( K \) at this temperature.
▶️Answer/Explanation
Convert units:
\( \Delta G^\circ = -34.2 \, \text{kJ/mol} = -34200 \, \text{J/mol} \)
Use the equation:
\( \Delta G^\circ = -RT \ln K \)
Rearranged: \( \ln K = -\frac{\Delta G^\circ}{RT} \)
\( \ln K = -\frac{-34200}{(8.314)(298)} = \frac{34200}{2477.572} \approx 13.80 \)
Exponentiate to solve for \( K \):
\( K = e^{13.80} \approx 9.91 \times 10^5 \)
Conclusion: Since \( K \gg 1 \), the reaction lies heavily toward the products at equilibrium.
Example
For the reaction:
\( \text{A}(g) + \text{B}(g) \rightleftharpoons \text{C}(g) \),the equilibrium constant at 298 K is \( K = 3.2 \times 10^{-4} \).
Calculate the standard Gibbs free energy change \( \Delta G^\circ \) in kJ/mol.
▶️Answer/Explanation
Use the equation:
\( \Delta G^\circ = -RT \ln K \)
Plug in values:
\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \), \( T = 298 \, \text{K} \), \( K = 3.2 \times 10^{-4} \)
\( \ln(3.2 \times 10^{-4}) \approx \ln(3.2) + \ln(10^{-4}) = 1.16 – 9.21 = -8.05 \)
(or use calculator directly)
\( \Delta G^\circ = -(8.314)(298)(\ln 3.2 \times 10^{-4}) = -(8.314)(298)(-8.05) \)
\( = +19897 \, \text{J/mol} = 19.9 \, \text{kJ/mol} \)
Conclusion: \( \Delta G^\circ = +19.9 \, \text{kJ/mol} \)
Since \( \Delta G^\circ > 0 \), the reaction is non-spontaneous under standard conditions and favors the reactants.
Example
For the reaction:
\( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \) the standard Gibbs free energy change at 298 K is \( \Delta G^\circ = -33.0 \, \text{kJ/mol} \).
At a certain point, the concentrations are:
[N2] = 0.50 mol/L, [H2] = 0.80 mol/L, [NH3] = 0.10 mol/L
Calculate \( \Delta G \) at this moment and determine whether the forward or reverse reaction is favored.
▶️Answer/Explanation
SWrite the expression for the reaction quotient \( Q \):
\( Q = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(0.10)^2}{(0.50)(0.80)^3} \)
\( Q = \frac{0.01}{0.50 \times 0.512} = \frac{0.01}{0.256} \approx 0.039 \)
Use the Gibbs equation:
\( \Delta G = \Delta G^\circ + RT \ln Q \)
Convert \( \Delta G^\circ \) to J/mol: \( -33.0 \, \text{kJ/mol} = -33000 \, \text{J/mol} \)
\( R = 8.314 \, \text{J/mol·K}, \, T = 298 \, \text{K} \)
\( \Delta G = -33000 + (8.314)(298) \ln(0.039) \)
\( \ln(0.039) \approx -3.24 \)
\( \Delta G = -33000 + (8.314)(298)(-3.24) \)
\( = -33000 – 8028 = -41028 \, \text{J/mol} = -41.0 \, \text{kJ/mol} \)
Conclusion: \( \Delta G = -41.0 \, \text{kJ/mol} \), which is more negative than \( \Delta G^\circ \)
The forward reaction is strongly favored under these non-standard conditions.